\(=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)

\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}\right)-2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

\(=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}\)

\(=\frac{2a}{\sqrt{a-4}}\) 

bài này không phải của lớp 7

Bài 1

a) \(P=\frac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)    (ĐK : x\(\ge0\) ; x\(\ne\) 1)

        \(=\frac{3a+\sqrt{9a}-3}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\)

         \(=\frac{3a+\sqrt{9a}-3-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{3a+\sqrt{9a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

         \(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)

b) \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}-1+2}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)

Vậy để P là số nguyên thì: \(\sqrt{a}-1\inƯ\left(2\right)\)

Mà Ư(2)={-1;1;2;-1}

=> \(\sqrt{a}-1\in\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

vậy a={0;4;9} thì P nguyên

Bài 2

  \(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)(ĐK:a\(\ge\)8)

      \(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

     \(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{1-\frac{4}{a}}\)

      \(=\sqrt{a-4}+2+\sqrt{a-4}-2:\frac{a-4}{a}\)

     \(=2\sqrt{a-4}\cdot\frac{a}{a-4}\)

     \(=\frac{2a}{\sqrt{a-4}}\)

a)    ĐK:  \(a\ge4\)

  \(P=\frac{\sqrt{a+4\sqrt{a-4}}+\sqrt{a-4\sqrt{a-4}}}{\sqrt{1-\frac{8}{a}+\frac{16}{a^2}}}\)

\(=\frac{\sqrt{\left(a-4\right)+4\sqrt{a-4}+4}+\sqrt{\left(a-4\right)-4\sqrt{a-4}+4}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

\(=\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{\left|1-\frac{4}{a}\right|}\)

\(=\frac{\sqrt{a-4}+2+\left|\sqrt{a-4}-2\right|}{1-\frac{4}{a}}\)

Nếu \(4\le a< 8\)thì:  \(P=\frac{\sqrt{a-4}+2+2-\sqrt{a-4}}{1-\frac{4}{a}}=\frac{4}{\frac{a-4}{a}}=\frac{4a}{a-4}\)

Nếu  \(a\ge8\)thì:  \(P=\frac{\sqrt{a-4}+2+\sqrt{a-4}-2}{1-\frac{4}{a}}=\frac{2\sqrt{a-4}}{\frac{a-4}{a}}=\frac{2a\sqrt{a-4}}{a-4}\)

\(P=\frac{\sqrt{\left(\sqrt{a-4}\right)^2+2.2.\sqrt{a-4}+4}+\sqrt{\left(\sqrt{a-4}\right)^2-2.2.\sqrt{a-4}+4}}{\sqrt{1^2-2.\frac{4}{a}}+\frac{4^2}{a^2}}\)

=\(\frac{\sqrt{\left(\sqrt{a-4}+2\right)^2}+\sqrt{\left(\sqrt{a-4}-2\right)^2}}{\sqrt{\left(1-\frac{4}{a}\right)^2}}\)

=\(\frac{|\sqrt{a-4}+2|+|\sqrt{a-4}-2|}{|1-\frac{4}{a}|}\)

=\(\frac{a-4+2+a-4-2}{1-\frac{4}{a}}\)

=\(\frac{2a-8}{\frac{a-4}{a}}\)

=\(\frac{2.\left(a-4\right)}{\frac{a-4}{a}}\)

=\(2.\left(a-4\right).\frac{a}{a-4}\)

=2a

(ĐKXĐ: a khác 4)

\(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)

\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\frac{4\sqrt{a}-4}{a-4}\)

\(=\frac{a+5\sqrt{a}+6-\left(a-3\sqrt{a}+2\right)-\left(4\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)

Bạn có thể đăng từng bài k như thế nhìn đã sợ ai làm

1)đặt nhân tử chung quy đồng là xong

2)phân tích x+2cănx-3=(1-cănx)(3+cănx)

3)2a+căn a đặt căn a ra r rút gọn